Question

Two vectors $A$ and $B$ have equal magnitudes. If magnitude of $A+B$ is equal to $n$ times the magnitude of $A-B$, then the angle between $A$ and $B$ is

• A. ${\cos }^{-1}\left(\frac{n-1}{n+1}\right)$
• B. ${\cos }^{-1}\left(\frac{n^2-1}{n^2+1}\right)$
• C. ${\sin }^{-1}\left(\frac{n-1}{n+1}\right)$
• D. ${\sin }^{-1}\left(\frac{n^2-1}{n^2+1}\right)$

Answer 1

Answer: (B)

Let $\theta$ be the angle between $A$ and $B$
$|A+B|=n|A-B|$
$\Rightarrow \sqrt{A^2+B^2+2AB\cos \theta }$
$=n\sqrt{A^2+B^2+2AB\cos (180-\theta )}$
$\Rightarrow |A|-|B|-A-B-x$
$\Rightarrow 2x^2(1+\cos \theta )=n^2\cdot 2x^2(1-\cos \theta )$
$1+\cos \theta =n^2-n^2\cos \theta$
$\left(1+n^2\right)\cos \theta =n^2-1$
$\cos \theta =\frac{n^2-1}{n^2+1}$
$\Rightarrow \theta ={\cos }^{-1}\left(\frac{n^2-1}{n^2+1}\right)$